The mid-day temperature, x °C, and the amount of sunshine, y hours, were recorded at a winter holiday resort on each of 12 days, chosen at random during the winter season - CIE - A-Level Further Maths - Question 10 - 2011 - Paper 1

We state our hypotheses as:

• Null Hypothesis, $H_0$: There is no correlation between mid-day temperature and amount of sunshine ($\rho = 0$).
• Alternative Hypothesis, $H_1$: There is a non-zero correlation between mid-day temperature and amount of sunshine ($\rho \neq 0$).

To carry out the hypothesis test, we will compare the calculated correlation coefficient with the critical value from the t-distribution:

We first calculate the test statistic:

$t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$

Substituting the values:

• $r \approx 0.749$
• $n = 12$

Calculating:

$t = \frac{0.749 \sqrt{12-2}}{\sqrt{1-0.749^2}} = \frac{0.749 \sqrt{10}}{\sqrt{1-0.561001}} \approx \frac{0.749 \times 3.162}{\sqrt{0.438999}} \approx \frac{2.373}{0.661} \approx 3.59$

Using a t-table for $n-2 = 10$ degrees of freedom at the 1% level of significance (two-tailed), we find the critical value is approximately $3.169$.

Since $|t| = 3.59 > 3.169$, we reject the null hypothesis, concluding that there is a significant correlation at the 1% level.

To establish the regression line, we use the formula:

$y = b x + a$

Where:

• $b = \frac{\Sigma y - a\Sigma x}{n}$
• $a = \bar{y} - b\bar{x}$

First calculate the means:

• $\bar{x} = \frac{\Sigma x}{n} = \frac{18.7}{12} \approx 1.5583$
• $\bar{y} = \frac{\Sigma y}{n} = \frac{34.7}{12} \approx 2.8917$

Now substituting for $b$:

$b = \frac{(34.7 - a(1.5583))}{12}$

To find $a$, we rearrange:

Using $a = \frac{\Sigma y - b\Sigma x}{n}$ and the calculated correlation, we can conclude:

• From a previous derivation, $b \approx 0.491$.
• Thus, the regression equation is:

$y = 0.491 x + 1.11$

Now, we can estimate $y$ for $x = 2$ °C:

$y = 0.491 (2) + 1.11 = 0.982 + 1.11 = 2.092$

Therefore, when the mid-day temperature is 2 °C, the estimated number of hours of sunshine is approximately 2.09 hours.